How do you use the sum or difference identities to find the exact value of #tan((23pi)/12)#?

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Answer 1 · 2016-11-30T12:43:25

#tan((23pi)/12) = frac (1 - sqrt(3)) (1+ sqrt(3)) #

First note that #(23pi)/12 =2pi-pi/12#

and as #tan (x+2pi) = tan(x)#, then:

#tan((23pi)/12) = tan(-pi/12)#

Now: #1/12=1/3-1/4#,

So:

#tan(-pi/12) = tan(pi/4-pi/3)#

Using:

#tan(alpha+beta) = frac (tan alpha - tan beta) (1+tan alpha tan beta)#

#tan((23pi)/12) = frac (tan (pi/4) - tan (pi/3)) (1+ tan (pi/4) tan (pi/3)) #

#tan((23pi)/12) = frac (1 - sqrt(3)) (1+ sqrt(3)) #